The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following formula:

$ \mathrm{KE} = \frac{1}{2} m v^2 = \frac{1}{2} m \frac{GM_e}{r} = \frac{GM_e m}{2r} $

For this satellite, the radius $ r $ is the sum of the Earth's radius $ R_E $ and the height $ h $ above the Earth's surface:

$ r = R_E + h $

Given:

Mass of the satellite, $ m = 1000 \, \text{kg} $

Mass of the Earth, $ M_e = 6 \times 10^{24} \, \text{kg} $

Radius of the Earth, $ R_E = 6.4 \times 10^6 \, \text{m} $

Height of orbit above the Earth's surface, $ h = 270 \, \text{km} = 2.7 \times 10^5 \, \text{m} $

Gravitational constant, $ G = 6.67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2} $

Substitute these values into the equation:

$ \mathrm{KE} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2(6.4 \times 10^6 + 2.7 \times 10^5)} $

$ = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2 \times 6.67 \times 10^6} $

$ = 3 \times 10^{10} \, \text{J} $

Hence, the kinetic energy of the satellite in its orbit is $ 3 \times 10^{10} \, \text{J} $.